Question

A 70 kg object strikes the ground with 2500 J of KE after falling freely from rest. How far above the ground was the object when it was released?

Answer 1

3.64 m

Step-by-step explanation:

m = Mass of object = 70 kg

Kinetic energy of the object = 2500 J

g = Acceleration due to gravity =
`9.81\ \text{m/s}^2`

h = Height from which the object is dropped

Kinetic energy is given by

`(1)/(2)mv^2=2500\\\Rightarrow v^2=(2* 2500)/(70)`

From conservation of energy we get kinetic energy equal to potential energy.

`(1)/(2)mv^2=mgh\\\Rightarrow h=(1)/(2g)v^2\\\Rightarrow h=(1)/(2* 9.81)* (2* 2500)/(70)\\\Rightarrow h=3.64\ \text{m}`

The object was released from a height of 3.64 m.